## anonymous 4 years ago A compound is 54.53% C, 9.15% H, and 36.32% O. What is its empirical formula? The molar mass of the compound is 132 amu. What is its molecular formula?

1. anonymous

i got it nevermind

2. JFraser

This requires a couple steps and a little imagination: assume you've got 100g of the substance. This turns the percentages into masses: 54.53g C, 9.15g H, 36.32g O Turn each mass into moles using the molar mass of each: $54.53g C * (\frac{1mol C}{12g C}) = 4.54mol C$$9.15g H * (\frac{1mol H}{1g H}) = 9.15mol H$$36.32g O * (\frac{1mol O}{16g O}) = 2.27mol O$ In order to turn each of these fractional moles into whole #s, divide them all by the smallest among them. In this case, 2.27 is the smallest: $\frac{4.54mol}{2.27mol} = 2mol C$$\frac{9.15mol}{2.27mol} = 4mol H$$\frac{2.27mol}{2.27mol} = 1mol O$ So your empirical formula is:$C{_2}H{_4}O$

3. anonymous

Thanks!